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Let $\mathbb{R}^\omega$ be endowed with the product topology. Is there a nonempty open set $U\subseteq \mathbb{R}^\omega$ such that $\mathbb{R}^\omega\cong \mathbb{R}^\omega\setminus \text{cl}(U)$?

(By $\text{cl}(\cdot)$ we denote the topological closure.)

Edit. Apologies for omitting the word "open" in the question above.

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    $\begingroup$ Taking $\mathbb R\times\mathbb R^\omega$ in place of $\mathbb R^\omega$ (as they are homeomorphic), won't $(-\infty,0]\times \mathbb R^\omega$ work? $\endgroup$
    – Wojowu
    Commented Jun 3, 2018 at 13:02
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    $\begingroup$ @Dominic: maybe you mean $U$ a compact subset? $\endgroup$
    – Pietro Majer
    Commented Jun 3, 2018 at 13:18
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    $\begingroup$ I'm voting to close this question as off-topic because -- in its original form at least -- it admits of a one line answer (as demonstrated in the comments). $\endgroup$
    – R.P.
    Commented Jun 3, 2018 at 13:25
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    $\begingroup$ Looking at the title rather than the question itself: is it obvious that $\mathbb R^\omega \setminus \{0\}$ is not homeomorphic to $\mathbb R^\omega $? $\endgroup$
    – Goldstern
    Commented Jun 3, 2018 at 13:45
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    $\begingroup$ @Goldstern You may want to ask this as a separate question, I would be interested to see an answer. $\endgroup$
    – Wojowu
    Commented Jun 3, 2018 at 15:50

1 Answer 1

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If $U= (-\infty,0) \times \mathbb R \times \mathbb R \times \dots$, then $\mathrm{cl}(U) = (-\infty,0] \times \mathbb R \times \mathbb R \times \dots$, and the complement of this is $(0,+\infty) \times \mathbb R \times \mathbb R \times \dots$, which is homeomorphic to the whole space.

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  • $\begingroup$ Wojowu beat me to it. $\endgroup$
    – Gerald Edgar
    Commented Jun 3, 2018 at 13:04
  • $\begingroup$ Edited, to get the changed question with $U$ open. $\endgroup$
    – Gerald Edgar
    Commented Jun 3, 2018 at 17:54

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